What is the integral from 0 to (2pi)/w of tsin(wt) ?

The integral from 0 to (2pi)/w of tsin(wt) is -(2pi)/(w^2)

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integral from 0 to (2pi)/w of tsin(wt)

\int_{0}^{\frac{2 π}{w}} t sin (wt) d t

- \frac{2 π}{w ^{2}}

Solution steps

\int_{0}^{\frac{2 π}{w}} t sin (wt) d t

Apply Integration By Parts

: [- \frac{1}{w} t cos (wt) - \int - \frac{1}{w} cos (wt) d t]_{0}^{\frac{2 π}{w}}

= [- \frac{1}{w} t cos (wt) - \int - \frac{1}{w} cos (wt) d t]_{0}^{\frac{2 π}{w}}

\int - \frac{1}{w} cos (wt) d t = - \frac{1}{w ^{2}} sin (wt)

= [- \frac{1}{w} t cos (wt) - (- \frac{1}{w ^{2}} sin (wt))]_{0}^{\frac{2 π}{w}}

Simplify

= [- \frac{1}{w} t cos (wt) + \frac{1}{w ^{2}} sin (wt)]_{0}^{\frac{2 π}{w}}

Compute the boundaries:

- \frac{2 π}{w ^{2}}

= - \frac{2 π}{w ^{2}}

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What is the integral from 0 to (2pi)/w of tsin(wt) ?
The integral from 0 to (2pi)/w of tsin(wt) is -(2pi)/(w^2)